Question

If the $K_b$ value in the hydrolysis reaction, $B^{+}+H_{2}O\rightleftharpoons BOH+H^{+}$ is $1.0\times {10}^{-6}$, then the hydrolysis constant of the salt would be:

• A. $1.0\times {10}^{-6}$
• B. $1.0\times {10}^{-7}$
• C. $1.0\times {10}^{-8}$
• D. $1.0\times {10}^{-9}$

Answer 1

Answer: (C)

$1.0\times {10}^{-8}$

$K_h=\frac{K_w}{K_b}=\frac{{10}^{-14}}{{10}^{-6}}={10}^{-8}$.